Question

An air-standard Stirling cycle operates with a maximum pressure of \(3600 \mathrm{kPa}\) and a minimum pressure of \(50 \mathrm{kPa}\) The maximum volume is 12 times the minimum volume, and the low-temperature reservoir is at \(20^{\circ} \mathrm{C}\). Allowing a \(5^{\circ} \mathrm{C}\) temperature difference between the external reservoirs and the air when appropriate, calculate the specific heat added to the cycle and its net specific work.

Short answer

In this air-standard Stirling cycle, the specific heat added to the cycle is zero due to the isothermal expansion process from state 1 to state 2. The net specific work of the cycle can be calculated using the formula: \(W_{net} = -\frac{C_v(T_3 - T_1)}{M_{air}}\), where \(T_3\) and \(T_1\) are the temperatures at state 3 and state 1, respectively, \(C_v\) is the specific heat capacity at constant volume, and \(M_{air}\) is the molar mass of air.

Step-by-step solution

Calculate the temperatures at all four points of the cycle

First, let's calculate the temperatures at all four points in the cycle: 1. At state 1, we have the lowest volume and lowest temperature. From the problem, we know that there is a 5-degree Celsius temperature difference between the air inside and the low-temperature reservoir, so \(T_1 = 20 + 5 = 25^{\circ}\mathrm{C} = 298\mathrm{K}\). 2. At state 2, we have the highest volume and lowest temperature. Since this is an isothermal expansion, the temperature stays the same, \(T_2 = T_1 = 298\mathrm{K}\). 3. At state 3, we have the maximum volume and highest temperature in the cycle. First, let's get the highest temperature. From the problem, we know that during the isothermal compression process, there is a 5-degree Celsius temperature difference between the air inside and the high-temperature reservoir. Let's denote the high-temperature reservoir as \(T_{3_r}\). So, \(T_3 = T_{3_r} - 5^{\circ} \mathrm{C}\). To find \(T_{3_r}\), we can use the thermodynamic relationship \(P_1V_1^\gamma=P_3V_3^\gamma\): $$ \frac{P_3}{P_1} = \left(\frac{V_1}{V_3}\right)^\gamma, \quad \text{where}\; V_3 = 12V_1 $$ 4. At state 4, we have the lowest volume and highest temperature in the cycle. Since this is an isothermal compression, the temperature stays the same, \(T_4 = T_3\).

Specific heat added during the isothermal expansion process

Now we'll find the specific heat added during the isothermal expansion process (from state 1 to state 2). The specific heat added (\(Q_1\)) can be calculated using the formula for isothermal processes: \(Q_1 = nC_v(T_2 - T_1)\), where \(C_v\) is the specific heat capacity at constant volume and \(n\) is the number of moles. Since \(T_1\) and \(T_2\) are the same (both are 298 K), the specific heat added during the isothermal expansion process is zero: \(Q_1 = 0\).

Calculate the net specific work of the cycle

Finally, we'll calculate the net specific work (\(W_{net}\)) for the Stirling cycle. We can use the following formula for the net work done on the cycle: \(W_{net} = Q_{23} + Q_{41}\), where \(Q_{23}\) is the heat transferred during the adiabatic compression process (state 2 to state 3) and \(Q_{41}\) is the heat transferred during the adiabatic expansion process (state 4 to state 1). For an ideal gas, we can express the heat transferred during adiabatic processes as follows: - \(Q_{23} = -nC_v(T_3 - T_2)\) - \(Q_{41} = -nC_v(T_4 - T_1)\) Since \(T_1 = T_2\) and \(T_3 = T_4\), we can rewrite the net work formula as: $$ W_{net} = -nC_v[(T_3 - T_2) + (T_4 - T_1)] = -nC_v[(T_3 - T_1) + (T_4 - T_2)] = -nC_v(T_3 - T_1) $$ So, the net specific work (\(W_{net}\)) for the Stirling cycle can be expressed as: \(W_{net} = -nC_v(T_3 - T_1)\). To make it specific work (work per unit mass), we need to divide it by the mass of air inside the cycle. Given the number of moles \(n\) and the molar mass of air \(M_{air}\), we can calculate the mass of air inside the cycle as follows: \(m_{air} = n \times M_{air}\). Hence, the specific work can be calculated as: $$ \frac{W_{net}}{m_{air}} = -\frac{nC_v(T_3 - T_1)}{n \times M_{air}} = -\frac{C_v(T_3 - T_1)}{M_{air}} $$ This is the formula we can use to calculate the specific work for the given Stirling cycle. Note that in this particular case, since \(Q_1 = 0\), the net specific work is equal to the specific heat added to the cycle.